Problem: Suppose we have a surface $S$ defined by the transformation $T$ for $\pi < u < 2\pi$ and $0 < v < \dfrac{\pi}{2}$. $T(u, v) = (2u, \cos(v) + \cos(u), v)$ What is the surface area of $S$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_0^{\frac{\pi}{2}} \int_{\pi}^{2\pi} 2\sqrt{2} \, du \, dv$ (Choice B) B $ \int_0^{\frac{\pi}{2}} \int_{\pi}^{2\pi} \sqrt{5 + 3\sin^2(v)} \, du \, dv$ (Choice C) C $ \int_0^{\frac{\pi}{2}} \int_{\pi}^{2\pi} \sqrt{4 + \sin^2(u) + 4\sin^2(v)} \, du \, dv$ (Choice D) D $ \int_0^{\frac{\pi}{2}} \int_{\pi}^{2\pi} \sqrt{4 + 5\sin^2(v)} \, du \, dv$
Assume we have a surface $S$ parameterized by a transformation $T$. If we want to find the surface integral over $S$ of a function $f$, we can use the formula below to convert it into a familiar double integral. $ \iint_S f(T(u, v)) | T_u \times T_v | \, du \, dv$ Finding surface area using a surface integral means using $f(x, y, z) = 1$. In effect, we are saying that we only care about the scaling factor caused by the area element. Therefore: $A = \int_0^{\frac{\pi}{2}} \int_{\pi}^{2\pi} |T_u \times T_v| \, du \, dv$ Now we need to find the magnitude of the area element. $|T_u \times T_v| = \sqrt{4 + \sin^2(u) + 4\sin^2(v)}$ [Calculation] In conclusion, the surface area of $S$ is: $ \int_0^{\frac{\pi}{2}} \int_{\pi}^{2\pi} \sqrt{4 + \sin^2(u) + 4\sin^2(v)} \, du \, dv$